# Is this function defined in terms of elliptic $\mathrm{K}$ integrals even?

Let $R,z > 0$ be positive real constants, and consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $$f(v) = \frac{1}{\sqrt{(R+v)^2+z^2}}\ \mathrm{K}\!\left( \frac{4 R v}{(R+v)^2+z^2} \right)$$ where $\mathrm{K}(m)$ is the complete elliptic integral of the first kind, defined (using Wolfram Mathematica's convention) by $$\mathrm{K}(m) = \int_0^{\pi/2} \frac{\mathrm{d} \theta}{\sqrt{1-m\sin^2 \theta}}.$$ Is $f$ an even function? Numerical tests suggest that it should be, but I can't find the right series of manipulations to demonstrate this analytically.

• Okay, backtracking somewhat, one obvious thing to do is divide everything by $R^2$, and then you can relabel $v/R \mapsto v$, $z/R \mapsto z$ to get rid of $R$. Not much progress, but at least that leaves you fewer variables! – Chappers Apr 29 '15 at 1:00
• Right, sorted it out (see below). – Chappers Apr 29 '15 at 1:24

Okay, cracked it. It's annoyingly simple in the end: $$K(m) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m\sin^2{\theta}}} \\ = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m(1-\cos^2{\theta})}} \\ = \int_0^{\pi/2} \frac{d\theta}{\sqrt{(1-m)+m\cos^2{\theta}}} \\ = \frac{1}{\sqrt{1-m}}\int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac{m}{m-1}\cos^2{\theta}}}$$ and changing variables you conclude that $$K(m) = \frac{1}{\sqrt{1-m}} K\left( \frac{m}{m-1} \right),$$ with caveats about square roots. Turns out you're fine, since $$1-m = 1- \frac{4v}{(1+v)^2+z^2} = \frac{(1+v)^2+z^2-4v}{(1+v)^2+z^2} = \frac{(1-v)^2+z^2}{(1+v)^2+z^2},$$ clearly positive. And then you also have $$\frac{m}{m-1} = -\frac{4v}{(1+v)^2+z^2} \frac{(1+v)^2+z^2}{(1-v)^2+z^2} = \frac{-4v}{(1-v)^2+z^2}$$
(Obviously I've set $R=1$ here, which is fine because I can just divide $v$ and $z$ by it.) Hence you end up with the evenness you asked for.
• That should be a $4Rv$ in the numerator of the argument of EllipticK, not $2Rv$. – David Zhang Apr 29 '15 at 0:17